CN104393579B - A Method to Overcome the Influence of Drain Current on Bus Differential Protection - Google Patents

Abstract

The invention provides a method for overcoming the influence of a drawing current on bus differential protection, which comprises the following steps: collecting and processing branch current signals; selecting a fault bus, and determining the branch current with the maximum amplitude in the branch circuits connected with the fault bus; and calculating the differential current and the braking current of the large difference element, and judging whether the large difference element acts or not. The method for overcoming the influence of the draw-out current on the bus differential protection provided by the invention does not need to reduce the braking coefficient during the split operation in a double-bus connection mode, can adaptively improve the sensitivity of the bus differential protection during the in-zone fault under the condition of the draw-out current, and simultaneously ensures the reliability during the out-zone fault.

Description

Translated fromChinese

一种克服汲出电流对母线差动保护影响的方法A Method to Overcome the Influence of Drain Current on Bus Differential Protection

技术领域technical field

本发明属于电力系统继电保护技术领域，具体涉及一种克服汲出电流对母线差动保护影响的方法。The invention belongs to the technical field of electric power system relay protection, and in particular relates to a method for overcoming the influence of drawn current on busbar differential protection.

背景技术Background technique

母线保护通常采用差动保护原理。差动保护由于其原理简单，不受振荡影响诸多优势得到了最广泛的应用。但是，在实际应用中母线差动保护的区内故障汲出电流问题，成为影响其安全性和可靠性的主要因素。本发明提出了母线差动保护克服汲出电流的影响的对策。Busbar protection usually adopts the differential protection principle. Differential protection is the most widely used due to its simple principle and many advantages of not being affected by oscillation. However, in practical applications, the current drawn by faults in the busbar differential protection zone has become the main factor affecting its safety and reliability. The invention proposes a countermeasure for overcoming the influence of the drawn current in the differential protection of the busbar.

对于双母线接线型式的母线保护，通常配置有大差和小差元件。大差保护用来判别是否在其保护范围内发生故障，而小差元件用来选择故障母线，并进行切除。双母线分列运行且两条母线通过周边电力网络形成电气连接时，当其中一条母线故障且另一条健全母线上存在供电电源。此供电电源向故障点提供的故障电流必然通过某一与非故障母线相连接的支路流出非故障母线，并通过与故障母线相连接的支路流向故障点，如说明附图1的即为汲出电流。对于常规比率差动算法，这个电流对大差的差动电流没有影响，却增大了制动电流，从而导致大差比率制动判据的灵敏度下降，在严重的情况下可导致由于大差保护拒动引起的整套母差保护拒动。因此，有的厂家的处理方式为在这种情况下，内部降低大差比率制动系数。对于双母双分段接线型式也存在类似的问题。For the busbar protection of the double busbar connection type, it is usually equipped with large difference and small difference elements. The large difference protection is used to judge whether a fault occurs within its protection range, while the small difference element is used to select the faulty bus and cut it off. When the double busbars run separately and the two busbars are electrically connected through the surrounding power network, when one of the busbars fails and there is a power supply on the other healthy busbar. The fault current provided by the power supply to the fault point must flow out of the non-fault bus through a branch connected to the non-fault bus, and flow to the fault point through a branch connected to the fault bus, as shown in Figure 1 is the current drawn. For the conventional ratio differential algorithm, this current has no effect on the differential current of the large difference, but increases the braking current, which leads to a decrease in the sensitivity of the braking criterion of the large difference ratio. The entire set of bus differential protection caused by the refusal to move. Therefore, some manufacturers deal with this situation by reducing the braking coefficient of the large difference ratio internally. Similar problems also exist for the double-female double-section wiring type.

发明内容Contents of the invention

为了克服上述现有技术的不足，本发明提供一种克服汲出电流对母线差动保护影响的方法，对于双母线接线方式无需在分列运行时降低制动系数，对于有汲出电流的情况可自适应地提高母差保护在区内故障时的灵敏度，同时保证区外故障时的可靠性。In order to overcome the deficiencies of the prior art above, the present invention provides a method for overcoming the influence of the drawn current on the differential protection of the busbar. For the double busbar connection mode, it is not necessary to reduce the braking coefficient during separate operation. Adaptively improve the sensitivity of the bus differential protection in the case of faults in the zone, and at the same time ensure the reliability in the case of faults outside the zone.

为了实现上述发明目的，本发明采取如下技术方案：In order to realize the above-mentioned purpose of the invention, the present invention takes the following technical solutions:

本发明提供一种克服汲出电流对母线差动保护影响的方法，所述方法包括以下步骤：The present invention provides a method for overcoming the influence of drawn current on bus differential protection, the method comprising the following steps:

步骤1：支路电流信号采集与处理；Step 1: Acquisition and processing of branch current signals;

步骤2：选取故障母线，并确定故障母线相连支路中幅值最大支路电流；Step 2: Select the faulty bus, and determine the maximum amplitude branch current in the branches connected to the faulty bus;

步骤3：计算大差元件的差动电流和制动电流，并判断大差元件是否动作。Step 3: Calculate the differential current and braking current of the large difference element, and judge whether the large difference element operates.

所述步骤1包括以下步骤：Described step 1 comprises the following steps:

步骤1-1：采集与母线连接的所有支路的电流采样值，并进行低通滤波，得到第j条支路第k次电流采样值i_j(k)，其中j＝1,2,…,n，n为与母线连接的支路总数；Step 1-1: Collect the current sampled values of all branches connected to the bus, and perform low-pass filtering to obtain the k-th current sampled value i_j (k) of the j-th branch, where j=1,2,… ,n, n is the total number of branches connected to the bus;

步骤1-2：对i_j(k)进行傅里叶变换得到第j条支路的电流相量的实部X_j和虚部Y_j，有：Step 1-2: Perform Fourier transform on i_j (k) to obtain the current phasor of the jth branch The real part X_j and imaginary part Y_j , have:

其中，N为基波在一个周期内的采样点数；Among them, N is the number of sampling points of the fundamental wave in one cycle;

再通过实步X_j和虚部Y_j得到的幅值I_jM和相角θ_j，有：Then through the real step X_j and the imaginary part Y_j to get The amplitude I_jM and phase angle θ_j , have:

所述步骤2包括以下步骤：Described step 2 comprises the following steps:

步骤2-1：计算小差元件的差动电流和制动电流；Step 2-1: Calculate the differential current and braking current of the small difference element;

小差元件的差动电流和制动电流分别用I_cd小和I_zd小表示，有：The differential current and braking current of the small difference element are expressed by I_{cd small} and I_{zd small} respectively, which are:

其中，m为与单段母线相连接的所有支路数；Among them, m is the number of all branches connected to the single-section bus;

步骤2-2：若某母线所对应的小差元件的差动电流和制动电流满足I_cd小＞k_res1I_zd小，则确定此母线为故障母线；其中k_res1为小差元件的比率制动系数，通常取为0.6；Step 2-2: If the differential current and braking current of the small difference element corresponding to a certain bus satisfy I_{cd small} > k_res1 I_{zd small} , then determine that this bus is a faulty bus; where k_res1 is the ratio of the small difference element The braking coefficient is usually taken as 0.6;

步骤2-3：选取确定的故障母线所连接支路中的幅值最大的支路电流Step 2-3: Select the branch current with the largest amplitude among the branches connected to the determined fault bus

所述步骤3包括以下步骤：Described step 3 comprises the following steps:

步骤3-1：计算大差元件差动电流，有：Step 3-1: Calculate the differential current of the large difference element, there are:

其中，I_cd为大差元件差动电流；Among them, I_cd is the differential current of the large difference element;

步骤3-2：计算大差元件的制动电流，有：Step 3-2: Calculate the braking current of the large difference element, there are:

其中，I_zd为大差元件的制动电流；为大差元件差动电流相量，且Among them, I_zd is the braking current of large difference element; is the differential current phasor of the large difference element, and

步骤3-3：判断大差元件是否动作，若满足比率制动判据I_cd＞k_resI_zd，即满足：Step 3-3: Judging whether the large difference element is activated, if the ratio braking criterion I_cd >k_res I_zd is satisfied, that is:

则表明大差元件动作，否则表明大差元件不动作，其中k_res为大差元件比率制动系数，取0.8。It indicates that the large difference element operates, otherwise it indicates that the large difference element does not act, where k_res is the ratio braking coefficient of the large difference element, which is 0.8.

与现有技术相比，本发明的有益效果在于：Compared with prior art, the beneficial effect of the present invention is:

1.在制动量的计算过程中，消除了汲出电流的影响。且由于与相位接近，其相量差的幅值远小于现有典型制动量，在差动量不变的情况下大幅提高了现有母线保护区内故障有汲出电流流出时大差元件的灵敏度；1. During the calculation of the braking amount, the influence of the drawn current is eliminated. and due to and The phases are close, and the magnitude of the phasor difference is much smaller than the existing typical braking amount, which greatly improves the sensitivity of the large difference element when the fault in the existing busbar protection zone has a drain current flowing out when the differential amount remains unchanged;

2.正常情况或区外故障时，为不平衡电流本发明所提出的判据演化为与常规判据相比，应用新型算法制动量相对于常规算法有所减小，但是由于为区外故障时的不平衡电流，在CT未饱和的情况下，的值非常小，母线差动保护仍能保证可靠不误动作；2. Under normal conditions or out-of-area faults, for the unbalanced current The criterion proposed by the present invention evolves as and conventional criteria Compared with the application of the new algorithm, the braking amount is reduced compared with the conventional algorithm, but due to is the unbalanced current when an external fault occurs, and when the CT is not saturated, The value of is very small, but the busbar differential protection can still guarantee reliable operation without malfunction;

3.通过与典型的母线电流差动判据比较可得，区内故障时，现有典型判据和本发明提出的判据动作量相同，本发明提出的判据制动量不受母线汲出电流的影响，且小于现有判据的制动量因此，本发明提出的判据灵敏度高于现有判据；区外故障时，本发明提出的判据与现有判据具有基本相同的可靠性。3. By the differential criterion with the typical bus current Comparably obtainable, when fault in the area, existing typical criterion and the criterion action amount that the present invention proposes Same, the criterion that the present invention proposes braking amount Not affected by the current drawn by the bus bar, and less than the braking amount of the existing criterion Therefore, the sensitivity of the criterion proposed by the present invention is higher than that of the existing criterion; when an out-of-area fault occurs, the criterion proposed by the present invention has basically the same reliability as the existing criterion.

附图说明Description of drawings

图1是现有技术中双母线接线区内故障存在汲出电流示意图；Fig. 1 is a schematic diagram of the current drawn by faults in the double-bus wiring area in the prior art;

图2是本发明实施例中克服汲出电流对母线差动保护影响的方法流程图。Fig. 2 is a flow chart of a method for overcoming the influence of drawn current on bus differential protection in an embodiment of the present invention.

具体实施方式detailed description

下面结合附图对本发明作进一步详细说明。The present invention will be described in further detail below in conjunction with the accompanying drawings.

如图2，本发明提供一种克服汲出电流对母线差动保护影响的方法，对于双母线接线方式无需在分列运行时降低制动系数，对于有汲出电流的情况可自适应地提高母差保护在区内故障时的灵敏度，同时保证区外故障时的可靠性。As shown in Figure 2, the present invention provides a method to overcome the influence of the drawn current on the differential protection of the busbars. For the double-busbar connection mode, it is not necessary to reduce the braking coefficient during separate operation, and the bus differential can be adaptively increased in the case of the drawn current. The sensitivity of the protection in the case of faults in the zone, while ensuring the reliability in the case of faults outside the zone.

克服汲出电流对母线差动保护影响的方法包括以下步骤：The method to overcome the effect of current draw on busbar differential protection includes the following steps:

步骤1：支路电流信号采集与处理；Step 1: Acquisition and processing of branch current signals;

步骤2：选取故障母线，并确定故障母线相连支路中幅值最大支路电流；Step 2: Select the faulty bus, and determine the maximum amplitude branch current in the branches connected to the faulty bus;

步骤3：计算大差元件的差动电流和制动电流，并判断大差元件是否动作。Step 3: Calculate the differential current and braking current of the large difference element, and judge whether the large difference element operates.

所述步骤1包括以下步骤：Described step 1 comprises the following steps:

步骤1-1：采集与母线连接的所有支路的电流采样值，并进行低通滤波，得到第j条支路第k次电流采样值i_j(k)，其中j＝1,2,…,n，n为与母线连接的支路总数；Step 1-1: Collect the current sampled values of all branches connected to the bus, and perform low-pass filtering to obtain the k-th current sampled value i_j (k) of the j-th branch, where j=1,2,… ,n, n is the total number of branches connected to the bus;

步骤1-2：对i_j(k)进行傅里叶变换得到第j条支路的电流相量的实部X_j和虚部Y_j，有：Step 1-2: Perform Fourier transform on i_j (k) to obtain the current phasor of the jth branch The real part X_j and imaginary part Y_j , have:

其中，N为基波在一个周期内的采样点数；Among them, N is the number of sampling points of the fundamental wave in one cycle;

再通过实步X_j和虚部Y_j得到的幅值I_jM和相角θ_j，有：Then through the real step X_j and the imaginary part Y_j to get The amplitude I_jM and phase angle θ_j , have:

所述步骤2包括以下步骤：Described step 2 comprises the following steps:

步骤2-1：计算小差元件的差动电流和制动电流；Step 2-1: Calculate the differential current and braking current of the small difference element;

小差元件的差动电流和制动电流分别用I_cd小和I_zd小表示，有：The differential current and braking current of the small difference element are expressed by I_{cd small} and I_{zd small} respectively, which are:

其中，m为与单段母线相连接的所有支路数；Among them, m is the number of all branches connected to the single-section bus;

步骤2-2：若某母线所对应的小差元件的差动电流和制动电流满足I_cd小＞k_res1I_zd小，则确定此母线为故障母线；其中k_res1为小差元件的比率制动系数，通常取为0.6；Step 2-2: If the differential current and braking current of the small difference element corresponding to a certain bus satisfy I_{cd small} > k_res1 I_{zd small} , then determine that this bus is a faulty bus; where k_res1 is the ratio of the small difference element The braking coefficient is usually taken as 0.6;

步骤2-3：选取确定的故障母线所连接支路中的幅值最大的支路电流Step 2-3: Select the branch current with the largest amplitude among the branches connected to the determined fault bus

所述步骤3包括以下步骤：Described step 3 comprises the following steps:

步骤3-1：计算大差元件差动电流，有：Step 3-1: Calculate the differential current of the large difference element, there are:

其中，I_cd为大差元件差动电流；Among them, I_cd is the differential current of the large difference element;

步骤3-2：计算大差元件的制动电流，有：Step 3-2: Calculate the braking current of the large difference element, there are:

其中，I_zd为大差元件的制动电流；为大差元件差动电流相量，且Among them, I_zd is the braking current of large difference element; is the differential current phasor of the large difference element, and

步骤3-3：判断大差元件是否动作，若满足比率制动判据I_cd＞k_resI_zd，即满足：Step 3-3: Judging whether the large difference element is activated, if the ratio braking criterion I_cd >k_res I_zd is satisfied, that is:

则表明大差元件动作，否则表明大差元件不动作，其中k_res为大差元件比率制动系数，取0.8。It indicates that the large difference element operates, otherwise it indicates that the large difference element does not act, where k_res is the ratio braking coefficient of the large difference element, which is 0.8.

最后应当说明的是：以上实施例仅用以说明本发明的技术方案而非对其限制，所属领域的普通技术人员参照上述实施例依然可以对本发明的具体实施方式进行修改或者等同替换，这些未脱离本发明精神和范围的任何修改或者等同替换，均在申请待批的本发明的权利要求保护范围之内。Finally, it should be noted that: the above embodiments are only used to illustrate the technical solutions of the present invention and not to limit them. Those of ordinary skill in the art can still modify or equivalently replace the specific implementation methods of the present invention with reference to the above embodiments. Any modifications or equivalent replacements departing from the spirit and scope of the present invention are within the protection scope of the claims of the pending application of the present invention.

Claims (1)

Translated fromChinese

1.一种克服汲出电流对母线差动保护影响的方法，其特征在于：所述方法包括以下步骤：1. A method for overcoming the influence of current drawn on busbar differential protection, characterized in that: the method may further comprise the steps:步骤1：支路电流信号采集与处理；Step 1: Acquisition and processing of branch current signals;步骤2：选取故障母线，并确定故障母线相连支路中幅值最大支路电流；Step 2: Select the faulty bus, and determine the maximum amplitude branch current in the branches connected to the faulty bus;步骤3：计算大差元件的差动电流和制动电流，并判断大差元件是否动作；Step 3: Calculate the differential current and braking current of the large difference element, and judge whether the large difference element operates;所述步骤1包括以下步骤：Described step 1 comprises the following steps:步骤1-1：采集与母线连接的所有支路的电流采样值，并进行低通滤波，得到第j条支路第k次电流采样值i_j(k)，其中j＝1,2,…,n，n为与母线连接的支路总数；Step 1-1: Collect the current sampled values of all branches connected to the bus, and perform low-pass filtering to obtain the k-th current sampled value i_j (k) of the j-th branch, where j=1,2,… ,n, n is the total number of branches connected to the bus;步骤1-2：对i_j(k)进行傅里叶变换得到第j条支路的电流相量的实部X_j和虚部Y_j，有：Step 1-2: Perform Fourier transform on i_j (k) to obtain the current phasor of the jth branch The real part X_j and imaginary part Y_j , have:${\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>=</span>=</span>\frac{\mathrm{<span><span>1</span>1</span>}}{\mathrm{<span><span>N</span>N</span>}}<span><span>\&lsqb;</span>\&lsqb;</span>\mathrm{<span><span>2</span>2</span>}\underset{\mathrm{<span><span>k</span>k</span>}<span><span>=</span>=</span>\mathrm{<span><span>1</span>1</span>}}{\overset{\mathrm{<span><span>N</span>N</span>}<span><span>-</span>-</span>\mathrm{<span><span>1</span>1</span>}}{<span><span>\&Sigma;</span>\&Sigma;</span>}}{\mathrm{<span><span>i</span>i</span>}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>(</span>(</span>\mathrm{<span><span>k</span>k</span>}<span><span>)</span>)</span>\mathrm{<span><span>s</span>the\; s</span>}\mathrm{<span><span>i</span>i</span>}\mathrm{<span><span>n</span>no</span>}<span><span>(</span>(</span>\mathrm{<span><span>k</span>k</span>}\frac{\mathrm{<span><span>2</span>2</span>}\mathrm{<span><span>\&pi;</span>\&pi;</span>}}{\mathrm{<span><span>N</span>N</span>}}<span><span>)</span>)</span><span><span>\&rsqb;</span>\&rsqb;</span>$${\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>=</span>=</span>\frac{\mathrm{<span><span>1</span>1</span>}}{\mathrm{<span><span>N</span>N</span>}}<span><span>\&lsqb;</span>\&lsqb;</span>\mathrm{<span><span>2</span>2</span>}\underset{\mathrm{<span><span>k</span>k</span>}<span><span>=</span>=</span>\mathrm{<span><span>1</span>1</span>}}{\overset{\mathrm{<span><span>N</span>N</span>}<span><span>-</span>-</span>\mathrm{<span><span>1</span>1</span>}}{<span><span>\&Sigma;</span>\&Sigma;</span>}}{\mathrm{<span><span>i</span>i</span>}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>(</span>(</span>\mathrm{<span><span>k</span>k</span>}<span><span>)</span>)</span>\mathrm{<span><span>c</span>c</span>}\mathrm{<span><span>o</span>o</span>}\mathrm{<span><span>s</span>the\; s</span>}<span><span>(</span>(</span>\mathrm{<span><span>k</span>k</span>}\frac{\mathrm{<span><span>2</span>2</span>}\mathrm{<span><span>\&pi;</span>\&pi;</span>}}{\mathrm{<span><span>N</span>N</span>}}<span><span>)</span>)</span><span><span>\&rsqb;</span>\&rsqb;</span>$其中，N为基波在一个周期内的采样点数；Among them, N is the number of sampling points of the fundamental wave in one cycle;再通过实部X_j和虚部Y_j得到的幅值I_jM和相角θ_j，有：Then get through real part X_j and imaginary part Y_j The amplitude I_jM and phase angle θ_j , have:${\mathrm{<span><span>I</span>I</span>}}_{\mathrm{<span><span>j</span>j</span>}\mathrm{<span><span>M</span>m</span>}}<span><span>=</span>=</span>\sqrt{\frac{{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>j</span>j</span>}}^{\mathrm{<span><span>2</span>2</span>}}<span><span>+</span>+</span>{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>j</span>j</span>}}^{\mathrm{<span><span>2</span>2</span>}}}{\mathrm{<span><span>2</span>2</span>}}}$${\mathrm{<span><span>\&theta;</span>\&theta;</span>}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>=</span>=</span>\mathrm{<span><span>a</span>a</span>}\mathrm{<span><span>r</span>r</span>}\mathrm{<span><span>c</span>c</span>}\mathrm{<span><span>t</span>t</span>}\mathrm{<span><span>g</span>g</span>}\frac{{\mathrm{<span><span>Y</span>Y</span>}}_{\mathrm{<span><span>j</span>j</span>}}}{{\mathrm{<span><span>X</span>x</span>}}_{\mathrm{<span><span>j</span>j</span>}}}$所述步骤2包括以下步骤：Described step 2 comprises the following steps:步骤2-1：计算小差元件的差动电流和制动电流；Step 2-1: Calculate the differential current and braking current of the small difference element;小差元件的差动电流和制动电流分别用I_cd小和I_zd小表示，有：The differential current and braking current of the small difference element are expressed by I_{cd small} and I_{zd small} respectively, which are:其中，m为与单段母线相连接的所有支路数；Among them, m is the number of all branches connected to the single-section bus;步骤2-2：若某母线所对应的小差元件的差动电流和制动电流满足I_cd小＞k_res1I_zd小，则确定此母线为故障母线；其中k_res1为小差元件的比率制动系数，取为0.6；Step 2-2: If the differential current and braking current of the small difference element corresponding to a certain bus satisfy I_{cd small} > k_res1 I_{zd small} , then determine that this bus is a faulty bus; where k_res1 is the ratio of the small difference element The braking coefficient is taken as 0.6;步骤2-3：选取确定的故障母线所连接支路中的幅值最大的支路电流Step 2-3: Select the branch current with the largest amplitude among the branches connected to the determined fault bus所述步骤3包括以下步骤：Described step 3 comprises the following steps:步骤3-1：计算大差元件差动电流，有：Step 3-1: Calculate the differential current of the large difference element, there are:${\mathrm{<span><span>I</span>I</span>}}_{\mathrm{<span><span>c</span>c</span>}\mathrm{<span><span>d</span>d</span>}}<span><span>=</span>=</span><span><span>|</span>|</span>\underset{\mathrm{<span><span>j</span>j</span>}<span><span>=</span>=</span>\mathrm{<span><span>1</span>1</span>}}{\overset{\mathrm{<span><span>n</span>no</span>}}{<span><span>\&Sigma;</span>\&Sigma;</span>}}{\stackrel{<span><span>\&CenterDot;</span>\&Center\; Dot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>|</span>|</span>$其中，I_cd为大差元件差动电流；Among them, I_cd is the differential current of the large difference element;步骤3-2：计算大差元件的制动电流，有：Step 3-2: Calculate the braking current of the large difference element, there are:${\mathrm{<span><span>I</span>I</span>}}_{\mathrm{<span><span>z</span>z</span>}\mathrm{<span><span>d</span>d</span>}}<span><span>=</span>=</span><span><span>|</span>|</span><span><span>(</span>(</span>{\stackrel{<span><span>\&CenterDot;</span>\&CenterDot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>c</span>c</span>}\mathrm{<span><span>d</span>d</span>}}<span><span>-</span>-</span>{\stackrel{<span><span>\&CenterDot;</span>\&Center\; Dot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>max</span>max</span>}}<span><span>)</span>)</span><span><span>-</span>-</span>{\stackrel{<span><span>\&CenterDot;</span>\&Center\; Dot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>m</span>m</span>}\mathrm{<span><span>a</span>a</span>}\mathrm{<span><span>x</span>x</span>}}<span><span>|</span>|</span>$其中，I_zd为大差元件的制动电流；为大差元件差动电流相量，且Among them, I_zd is the braking current of large difference element; is the differential current phasor of the large difference element, and步骤3-3：判断大差元件是否动作，若满足比率制动判据I_cd＞k_resI_zd，即满足：Step 3-3: Judging whether the large difference element is activated, if the ratio braking criterion I_cd >k_res I_zd is satisfied, that is:$<span><span>|</span>|</span>\underset{\mathrm{<span><span>j</span>j</span>}<span><span>=</span>=</span>\mathrm{<span><span>1</span>1</span>}}{\overset{\mathrm{<span><span>n</span>no</span>}}{<span><span>\&Sigma;</span>\&Sigma;</span>}}{\stackrel{<span><span>\&CenterDot;</span>\&CenterDot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>j</span>j</span>}}<span><span>|</span>|</span><span><span>></span>></span>{\mathrm{<span><span>k</span>k</span>}}_{\mathrm{<span><span>r</span>r</span>}\mathrm{<span><span>e</span>e</span>}\mathrm{<span><span>s</span>the\; s</span>}}<span><span>|</span>|</span><span><span>(</span>(</span>{\stackrel{<span><span>\&CenterDot;</span>\&CenterDot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>c</span>c</span>}\mathrm{<span><span>d</span>d</span>}}<span><span>-</span>-</span>{\stackrel{<span><span>\&CenterDot;</span>\&Center\; Dot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>m</span>m</span>}\mathrm{<span><span>a</span>a</span>}\mathrm{<span><span>x</span>x</span>}}<span><span>)</span>)</span><span><span>-</span>-</span>{\stackrel{<span><span>\&CenterDot;</span>\&CenterDot;</span>}{\mathrm{<span><span>I</span>I</span>}}}_{\mathrm{<span><span>m</span>m</span>}\mathrm{<span><span>a</span>a</span>}\mathrm{<span><span>x</span>x</span>}}<span><span>|</span>|</span>$则表明大差元件动作，否则表明大差元件不动作，其中k_res为大差元件比率制动系数，取0.8。It indicates that the large difference element operates, otherwise it indicates that the large difference element does not act, where k_res is the ratio braking coefficient of the large difference element, which is 0.8.